Question

please solve for X. Thanks​

Answer 1

Answer:

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Answer 2

$\dfrac{1}{a+b+x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x}\\\\\dfrac{1}{a+b+x}-\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b}\\\\\dfrac{x}{x(a+b+x)}-\dfrac{a+b+x}{x(a+b+x)}=\dfrac{b}{ab}+\dfrac{a}{ab}\\\\\dfrac{x-a-b-x}{x(a+b+x)}=\dfrac{a+b}{ab}\\\\\dfrac{-a-b}{x(a+b+x)}=\dfrac{a+b}{ab}\\\\\dfrac{a+b}{-x(a+b+x)}=\dfrac{a+b}{ab}\\\\-x(a+b+x)=ab\\-ax-bx-x^2=ab\\x^2+ax+bx+ab=0\\x(x+a)+b(x+a)=0\\(x+b)(x+a)=0\\x=-b \vee x=-a$